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Vector geometry: Proof of a trapezium and cross and dot product help

  1. Nov 3, 2012 #1
    1. The problem statement, all variables and given/known data
    #1 Given that the angle between the vectors a and b is 2Pi/3 and |a|=3 and |b|=4 calculate:
    (axb)^2 [(2a+b)x(a+2b)]^2

    #2 Given three unit vectors, a, b, c such that a+b+c=0 find (a dot b) + (b dot c) + (c dot a)

    #3 Given AB=a+2b BC=-4a-b CD= -5a-3b where a and b are any two vectors, prove that ABCD is a trapezium


    2. Relevant equations

    Given above

    3. The attempt at a solution
    #1 I know that axb = |a||b|sinx
    and axa = 0 and axb = -b x a but i do not understand how the powers or constants affect this?

    #2 No idea

    #3 A trapezium has one set of parallel sides only and i know if vectors are parallel they are a multiple of each other, however none of these seem to be multiples of each other?
     
  2. jcsd
  3. Nov 3, 2012 #2

    micromass

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    This is false. The length of [itex]a\times b[/itex] equals [itex]|a||b|\sin\theta[/itex].

    Anyway, what do you mean with [itex](a\times b)^2[/itex]?? What does it mean to take the square of a vector??
     
  4. Nov 3, 2012 #3
    I am using x as the cross product, meaning the cross product of a and b is |a||b|sinx, is this not correct?
     
  5. Nov 3, 2012 #4
    and (a x b)^2 means the cross product of a and b all sqaured
     
  6. Nov 3, 2012 #5

    micromass

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    No, that is not correct. Review your definition of cross product.
     
  7. Nov 3, 2012 #6

    micromass

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    Yes, I get that: it means that you take the square of the cross product. But the cross product is a vector. What do you mean with the square of a vector?
     
  8. Nov 3, 2012 #7
    This is the definition of cross product online |a x b|=|a||b|sinx
    I have no idea what you mean vectors is a very confusing topic for me
     
  9. Nov 3, 2012 #8

    micromass

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    Yes, the definition [itex]|a\times b|=|a||b|\sin x[/itex] is correct. But writing [itex]a\times b=|a||b|sin x[/itex] is not correct. Do you see the difference?
     
  10. Nov 3, 2012 #9
    Oh i see, so how does the formula differ without the modulus sign? and what happens where there is addition involved as in Q1 part 2
     
  11. Nov 3, 2012 #10

    micromass

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    Do you understand that [itex]a\times b[/itex] is a vector and that [itex]|a\times b|[/itex] is a scalar??
     
  12. Nov 3, 2012 #11
    yes i do now, but given this information how would i write the vector a x b?
     
  13. Nov 3, 2012 #12

    micromass

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    Before you do that, you got to understand what [itex](a\times b)^2[/itex] is. What does it mean to square a vector?
     
  14. Nov 3, 2012 #13
    Is it the dot product of the two vectors, creating a scalar?
     
  15. Nov 3, 2012 #14

    micromass

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    Yes.

    So you need to find [itex](a\times b)\cdot (a\times b)[/itex].

    Given a vector v, what is the definition of v.v?
     
  16. Nov 3, 2012 #15
    |v|^2?
     
  17. Nov 3, 2012 #16

    micromass

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    So [itex](a\times b)\cdot (a\times b)=|a\times b|^2[/itex]. Now you should be able to calculate this.
     
  18. Nov 3, 2012 #17
    Thankyou! I have got the answer 108?

    So for [(2a+b) x(a+2b)]^2 I do [(2a+b)x(a+2b)] dot [(2a+b)x(a+2b)] = |(2a+b)x(a+2b)|^2

    I know to do this i need to use the distributive law. Is it

    2a x a + 2a x2b + b xa + b x2b?
    Where 2a x a = 0 and bx2b=0?
     
  19. Nov 3, 2012 #18
    So the answer is 108? Thankyou!
    For [(2a+b) x(a+2b)]^2 is this the same as
    (2a+b) x(a+2b) dot (2a+b) x(a+2b) = |(2a+b) x(a+2b)|^2 ??

    I know i have to use some sort of distributive law here but i am not sure on how to do it would it be 2a x a + 2a x 2b + b x a + b x 2b ? where 2a x a = 0 and b x 2b =0??
     
  20. Nov 3, 2012 #19
    Sorry didnt think the first message had sent!
     
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